A metal wire of resistance $3\,\Omega $ is elongated to make a uniform wire of double its previous length. The new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle $60^o$ at the centre, the equivalent resistance between these two points will be
JEE MAIN 2019, Medium
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$R=\frac{\rho \ell}{A}=\frac{\rho \ell}{(V / \ell)}=\frac{\rho \ell^{2}}{V} \quad(V \rightarrow \text { Volume of wire })$
$\Rightarrow$ Final resistance $=3 \times(\mathrm{B})^{2}=12 \,\Omega$
$\begin{aligned} \mathrm{R}_{\mathrm{eq}} &=2\, \Omega \| 10\, \Omega \\ &=\frac{5}{3} \,\Omega \end{aligned}$
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